# Tag Info

15

You can use a high vertical tube to store water in it (fill it from the bottom by pushing the water in) How much water can you store? It obviously depends on the pressure you apply to push it in. If you push harder, there will be more water stored. The tube is characterized not the amount of water, but by how easy it is to store the water. Its "capacity" ...

4

Capacitance is "charge over voltage" – and one farad is "coulomb per volt" – because the capacity of capacitors (something that determines their "quality") is the ability to store a maximum charge on the plate ($+Q$ on one side, $-Q$ on the other side) given a fixed voltage. When you try to separate the charges, you unavoidably create electric fields ...

3

While the other two answers are technically correct, they are not actually addressing the engineering aspects of what comprises a "good" capacitor. Ideal capacitors in parallel or series circuits lead to ideal capacitors of different value, i.e. there is no measurable "quality" difference - ideal is ideal. In reality, of course, there are no ideal ...

3

First answer this: does it take more power to run one light bulb or two? Transmission lines are designed to be low loss, but they run a long way. Lightbulbs are designed to be "lossy" because that's how they work. Let's say for the sake of argument that the whole transmission circuit loses 100 watts and you're talking about a 100 watt bulb: those numbers ...

3

The curves show a charging that is proportional to $1-\mathrm{exp}(-t/\tau)$. Essentially, you should flip the exponential decay graph upside down.

2

If you have a known load $R'$, you can divide the potential of a source with a single resistor. If your source voltage is $V$ and you want a voltage $V'$ across the load, you can write $$V' = \frac{R'}{R_L+R'}V$$ from which you solve for $R_L$, the load. However, if you don't know $R'$ (or if it can change - for example, as a light bulb heats up, its ...

2

if we connect the set up like this the bulb glows If the pin positions in the photo are Ground Neutral Line And you are in a country where the neutral is linked to ground at the main panel but where protective devices such as GFCI or RCD are not used. Then you have a path for current to flow from Line via Ground to Neutral. if ...

2

(1) If the wave is induced by and propagation from the voltage source (battery), then it should take the vector path of the magnetic field created by the battery, instead of the circuit path. Yes, and the magnetic field follows the circuit path, but remains outside the wires. The electric current in a wire is causing a magnetic field which is located ...

2

For example in simple circuit with lamp. I understand that the energy is spent for heating and lighting. But how exactly it happens? The MIT Teal project has an excellent short video which can be used as an animated diagram of the (Classical physics) process inside a resistor such as a tungsten filament. In the video below, imagine that the vertical ...

2

The only lossy component in the circuit is the resistor and the power dissipated in the resistor is $I_{\text{rms}} R$ so the conversion $I_{\text{rms}} = \dfrac {I_{\text{peak}} }{\sqrt{2}}$ has to be made assuming that the variations of current and voltage are sinusoidal. The mean power dissipated in the capacitor over a cycle is zero.

2

To this EE's eyes, both formulas are incorrect; the reactance of a capacitor is $$X_C = -\frac{1}{\omega C}$$ and so the sign is incorrect in the first formula. Since reactance is the imaginary part of impedance $$Z = R + iX$$ reactance is a real number and so the second formula can't be correct. It is instead the formula for the impedance of a ...

2

Note that the sum of two phasors is another phasor: $Ae^{j(wt+\phi_1)} + Be^{j(wt+\phi_2)} = Ce^{j(wt+\phi_3)}$ Where A,B,C are real. The only way for the real part of the right side to be equal to zero at all times is if $C=0$. In which case the whole thing is $0$, both the real and complex parts are 0. So a sum of phasors (of the full sinusoidal form ...

2

The current doesnt pass through the voltmeter so the the resistance of load R is not affected.

2

The idea of the linked page is this: you can often find the equivalent resistance of certain highly symmetrical assemblies of resistors by adding "phantom" wires between points with equal potential. This has the advantage of giving quick, near-automatic answers without the use of Kirchhoff's laws, but it is very limited and works in very few cases. This ...

2

It is sometimes easier to visualise what is happening by using the idea of potential. To do this make one point in the circuit 0 V. This is a totally arbitrary choice. It is the bottom right hand corner of your circuit. Note to make the sums easier I have change the emf of the battery to 90 V so 2 A flows through the battery and 1 A through each of the ...

2

When you apply an AC voltage to a capacitor, current will flow into the capacitor, and back out again. As long as the absolute voltage on the AC generator is higher than on the capacitor, current will flow to increase the charge on the capacitor; and when it's smaller, current will flow to decrease the charge. Note that since there is an equal and opposite ...

2

For copper the temperature coefficient of resistivity is $3.9\times 10^{-3} \text{K}^{-1}$ and the temperature coefficient of thermal linear expansion is $1.6\times 10^{-4} \text{K}^{-1}$. They differ by a factor of about 24 so a change in temperature will cause a bigger change in resistance than in the linear dimensions of copper. Resistance is given by ...

2

Georg Ohm's original experiments, 1825, established that for a set temperature, the current through a specific length of a conductor was proportional to the potential difference applied. Ohm's law is empirical; it cannot be derived directly from Maxwell's equations as it depends upon material properties. It is violated by many materials, and even then ...

2

Assuming the variable resistor and the wire are connected in series you can find their total equivalent resistance. From there you can calculate the current through them and then, knowing current and resistance, you can find the voltage drop across the wire.

2

I believe the main effect of Earth Hour is in attracting public attention to sustainability. But besides of it, without doubt it saves energy (... and natural resources as well as one's wallet) if one turns off their unused light. The scale of one hour savings on turning off dispensable lighting for one hour is however rather small compared to other means ...

2

We Use $C=Q/V$ because those were useful things to measure. It's often easy to forget, but many of the equations we use are chosen because the work, and because other equations didn't work. Never underestimate that part of the reality. We don't use "charge per unit volume" because that number is not constant. You can charge a capacitor up without ...

2

I understand that capacitance is the ability of a body to store an electrical charge and the formula is $C = {Q \over V}$ Perhaps you just need to top thinking of capacitance as that. "Capacitance" sounds like "capacity", which leads to an intuitive trap like this: If I have a basket with a capacity of 2 apples, then a basket with more capacity can ...

2

A capacitor is used to store energy in form of electric fields. This electric field is created by charges on plates of capacitor. So, basically you are storing charge on capacitors. Let someone ask you how much charge you can store in your capacitor.What would you reply? Clearly , you reply " I may store 1mC or 100mC, depending on Potential difference ...

1

For instance, why don't measure the ability to store something by the volume it takes so why not charge per unit volume. There is nothing wrong with you defining a parameter which is the "charge per unit volume" but after defining it then what are you going to do with it? So here you have a capacitor and its charge per unit volume is $3 \;\text{C ... 1 I want to charge a 12v 100ah battery which need 20 amp of current. You can charge the battery at any current proving it is not too high and it might be that the 20 amp is the maximum charging current? If the adapter gives a constant (regulated) 12 volts then you will not be able to charge a battery of the lead-acid type it will require more than 12 V ... 1 If the inner conducting shell is charged then the charges will reside on the outside of the inner shell as there can be no electric field inside a conducting shell. The charges on the outside of the inner conducting shell will produce a radial electric field and the outer conducting shell will find itself in that electric field. However the outer ... 1 As has already been pointed out there is no quick fix in that you have to solve some simultaneous equations which you have obtained using nodes or loops and then it might be easier to solve the simultaneous equations using matrices? Possibly you might find it easier to solve this problem using superposition? 1 I'm not sure exactly what you're asking for. But let's say there's an electric current flowing through a straight wire segment of length$l$, then the change in$\Delta\phi$, or$V$, would be defined by $$\Delta\phi = \int \mathbf{E}\cdot d\mathbf{l}$$ Because it is a straight wire, $$\Delta\phi = E\cdot l$$ But we have a definition of current$$I = ... 1 why would increasing voltage, while keeping charge constant, have any effect on the ability of a body to store charge. (1) Capacitors don't store charge, they store electrical energy. For a capacitor, it is understood that one plate has charge$Q$while the other plate has charge$-Q\$ so there is no net electric charge stored. (2) If you increase ...

1

in terms of making a better capacitor that can store more charge would you use in series or in parallel? To be sure, capacitors don't (ordinarily) store charge, capacitors store energy, i.e., a 'charged' capacitor is electrically neutral. If, by better, you mean store more energy for a given voltage, then you want the combination of capacitors to be ...

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